Seb G. 11 Plus Maths tutor, 13 plus  Maths tutor, A Level Maths tutor...

Seb G.

£18 - £20 /hr

Mathematics (Masters) - Bath University

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About me

About Me:

I'm a third-year maths student at the University of Bath. Having had a rocky start with the subject, I didn't fall in love with it until later on so I can resonate with the less enthused students. Having done maths and further maths A-levels, I now specialise in pure maths so that's where my strengths lie. I am most adept at tutoring the Core and Further Pure units, although I can also tutor every other area barring statistics (sadly I never studied any statistics as it didn't interest me as much as the other unit choices).


Prior to Tutoring:

Before meeting me, in your initial message, could you please include the following pieces of information:


Level of tutoring (e.g. GCSE, A-level, general secondary school)Any particular exams you're working towards, plus the exam boardsYour favourite joke (this won't matter, but it might amuse me)

At this point in time, I am undergoing the biannual exam season. I shall endeavor to reply to any messages as quickly as possible, but please be patient with me! :)

About Me:

I'm a third-year maths student at the University of Bath. Having had a rocky start with the subject, I didn't fall in love with it until later on so I can resonate with the less enthused students. Having done maths and further maths A-levels, I now specialise in pure maths so that's where my strengths lie. I am most adept at tutoring the Core and Further Pure units, although I can also tutor every other area barring statistics (sadly I never studied any statistics as it didn't interest me as much as the other unit choices).


Prior to Tutoring:

Before meeting me, in your initial message, could you please include the following pieces of information:


Level of tutoring (e.g. GCSE, A-level, general secondary school)Any particular exams you're working towards, plus the exam boardsYour favourite joke (this won't matter, but it might amuse me)

At this point in time, I am undergoing the biannual exam season. I shall endeavor to reply to any messages as quickly as possible, but please be patient with me! :)

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About my sessions

In most cases, we'll cover what you request. In some cases, however, we'll need to cover some background material prior to the requested topics. This is to aid your understanding and, whilst it might seem tedious at first, missing this knowledge out could result in even more confusion than before. Maths is like a large canvas that's slowly being painted. As you understand a certain area of maths, you paint that area on the canvas. If, however, you try to jump to something without covering the necessary background material, you might end up with something that doesn't fit with the rest of the painting. It is essential to work towards a goal rather than jumping in at the deep end. I'll use whatever methods work for you, and help you understand it well enough to explain it back to me or your classmates. To quote Peter Singer, "whatever cannot be said clearly is probably not being thought clearly either".

In most cases, we'll cover what you request. In some cases, however, we'll need to cover some background material prior to the requested topics. This is to aid your understanding and, whilst it might seem tedious at first, missing this knowledge out could result in even more confusion than before. Maths is like a large canvas that's slowly being painted. As you understand a certain area of maths, you paint that area on the canvas. If, however, you try to jump to something without covering the necessary background material, you might end up with something that doesn't fit with the rest of the painting. It is essential to work towards a goal rather than jumping in at the deep end. I'll use whatever methods work for you, and help you understand it well enough to explain it back to me or your classmates. To quote Peter Singer, "whatever cannot be said clearly is probably not being thought clearly either".

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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06/02/2018

Qualifications

SubjectQualificationGrade
MathsA-level (A2)A*
Further MathsA-level (A2)A
PhysicsA-level (A2)A
GeologyA-level (A2)B

General Availability

Pre 12pm12-5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
MathsGCSE£18 /hr
Maths13 Plus£18 /hr
Maths11 Plus£18 /hr

Questions Seb has answered

How do I differentiate implicitly?

The most important thing to remember when differentiating implicitly is that y is a function of x. Rewriting y as y(x) often makes it much clearer. For example, evaluate d/dx (y2): using the aforementioned notation, this becomes d/dx [y(x)]2. By the chain rule, it is easy to see that this is equal to dy/dx * 2y.

Perhaps an easier way of remembering this is to differentiate with respect to y, then multiply by dy/dx. For example, evaluate d/dx(ln(y)): to find the answer, we differentiate ln(y) with respect to y to get 1/y, then multiply this by dy/dx to get dy/dx * 1/y.

The above method works because of the chain rule, which states that df/dx = df/dy * dy/dx. All we are doing is renaming the function as f (in the first example f = y2, in the second example f = ln(y)) and applying this result.

The most important thing to remember when differentiating implicitly is that y is a function of x. Rewriting y as y(x) often makes it much clearer. For example, evaluate d/dx (y2): using the aforementioned notation, this becomes d/dx [y(x)]2. By the chain rule, it is easy to see that this is equal to dy/dx * 2y.

Perhaps an easier way of remembering this is to differentiate with respect to y, then multiply by dy/dx. For example, evaluate d/dx(ln(y)): to find the answer, we differentiate ln(y) with respect to y to get 1/y, then multiply this by dy/dx to get dy/dx * 1/y.

The above method works because of the chain rule, which states that df/dx = df/dy * dy/dx. All we are doing is renaming the function as f (in the first example f = y2, in the second example f = ln(y)) and applying this result.

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2 years ago

599 views

How do I evaluate composite functions?

Suppose you have 2 functions: f(x) = 3x2, g(x) = log3(x). These are arbitrary, any functions would work. Evaluate f(g(x)): let y = log3(x) ( = g(x) ), then f(g(x)) = f(y) = 3y2 = 3[log3(x)]2.

The part that people tend to find difficult is remembering what it means to apply a function. A simple subsitution makes this much easier. Whilst the above situation makes it seem easy, consider how much more confusing it could be if f(x) = [x7 + 9x5 + e5x + cos(x-1/3)]/[sin(ex/6) + 1729*x], and g(x) was something similarly complicated; a simple substitution can do wonders and will help prevent confusion.

Suppose you have 2 functions: f(x) = 3x2, g(x) = log3(x). These are arbitrary, any functions would work. Evaluate f(g(x)): let y = log3(x) ( = g(x) ), then f(g(x)) = f(y) = 3y2 = 3[log3(x)]2.

The part that people tend to find difficult is remembering what it means to apply a function. A simple subsitution makes this much easier. Whilst the above situation makes it seem easy, consider how much more confusing it could be if f(x) = [x7 + 9x5 + e5x + cos(x-1/3)]/[sin(ex/6) + 1729*x], and g(x) was something similarly complicated; a simple substitution can do wonders and will help prevent confusion.

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2 years ago

590 views

How do I find the prime factors of a number?

The most common way of doing this is called a bubble search (this may not be the technical name, but it's the one I was taught). The method is as follows:

Start by writing the number at the top of your page. Now find 2 numbers which multiply to that number (e.g. starting with 100, I might find 2 and 50). Draw lines down from 100 to the numbers 2 and 50. Check whether or not these numbers are primes (e.g. check whether 2 and 50 are primes - 2 is, 50 isn't). If you have found a prime number, draw a bubble around it. Now do the same process for 50. I find the factors 2 and 25. 2 is prime, 25 isn't. I draw a bubble around 2 and start again from 25. I find the numbers 5 and 5. both are prime so I bubble both.

Now that I've circled both numbers, my search is complete: the prime factors are the numbers in bubbles.

Note that, if neither number is prime, you have to apply the method to both numbers. Because of this, it's often best to find a prime number for one of the 2 numbers.

Some tricks to make this easier are:

If the number is even, use 2 as one of your factors.

If the number ends with a 5 or a 0, use 5 as one of your factors.

If the digits of the number add up to a multiple of 3, use 3 as one of your factors.

The most common way of doing this is called a bubble search (this may not be the technical name, but it's the one I was taught). The method is as follows:

Start by writing the number at the top of your page. Now find 2 numbers which multiply to that number (e.g. starting with 100, I might find 2 and 50). Draw lines down from 100 to the numbers 2 and 50. Check whether or not these numbers are primes (e.g. check whether 2 and 50 are primes - 2 is, 50 isn't). If you have found a prime number, draw a bubble around it. Now do the same process for 50. I find the factors 2 and 25. 2 is prime, 25 isn't. I draw a bubble around 2 and start again from 25. I find the numbers 5 and 5. both are prime so I bubble both.

Now that I've circled both numbers, my search is complete: the prime factors are the numbers in bubbles.

Note that, if neither number is prime, you have to apply the method to both numbers. Because of this, it's often best to find a prime number for one of the 2 numbers.

Some tricks to make this easier are:

If the number is even, use 2 as one of your factors.

If the number ends with a 5 or a 0, use 5 as one of your factors.

If the digits of the number add up to a multiple of 3, use 3 as one of your factors.

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2 years ago

669 views

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